Your question reaches down to the foundation of set theory. A common approach is Zermelo–Fraenkel set theory ZF. It does not define a set as a "collection of objects" or something like that (although that is a nice intuitive understanding), but gives a list of axioms which have to be satisfied by things called sets.
However, it turns out that some "collections" of objects which one can easily name, for example
the collection of all sets,
the collection of all compact metric spaces,
cannot be sets (if they were sets, we would get contradictions as Russell's paradox). Such collections are called proper classes, and in a sense they are too big to satisfy all axioms of ZF. For example, there is no set containing as elements all sets and no set of all compact metric spaces.
Concerning your questions:
The compact metric spaces do not form a set. This is not a drama, because it is possible to find a set of representatives. If you are satisfied with compact metrizable toplogical spaces (i.e. if you are not interested in specific metrics because all metrics generating the same compact topology on a given set are uniformly equivalent), then Daniel Wainfleet's comment shows what to do: The closed subspaces of the Hilbert cube $Q$ form a set $\mathfrak C$. All $X \in \mathfrak C$ are compact metrizable and each compact metrizable space is homeomorphic to some member of $\mathfrak C$. If you want to consider compact metric spaces, then it is just a little more complicated. The previous argument shows that each compact metric space $(X,d)$ admits a bijection $\phi : X \to X'$ to a subset $X'$ of $Q$. This $\phi$ induces a metric $d_\phi$ on $X'$ making $(X,d)$ and $(X',d_\phi)$ isometrically isomorphic. The metrics on a given set $X'$ form a set, thus 2. shows that all pairs $(X',d')$ where $X'$ is a subset of the Hilbert cube (i.e. a member of the power set of $Q$) and $d'$ is a metric on $X'$ also form a set $\mathfrak M$. Not all such $(X',d')$ are compact, but the subset $\mathfrak M_c$ of compact $(X',d') \in \mathfrak M$ gives you the desired set of representatives for the class of compact metric spaces.
Yes. As Berci comments, this is the replacement axiom of ZF.